# 14.1: C.1- Infintesimal Rotations

Any rotation matrix can be written as the identity plus something:

$\underset{\sim}{C}=\underset{\sim}{\delta}+\underset{\sim}{r}$

The orthogonality requirement $$\underset{\sim}{C}^T \underset{\sim}{C} = \underset{\sim}{\delta}$$ can then be expressed as:

\begin{aligned} \underset{\sim}{C}^{T} \underset{\sim}{C} &=\left(\underset{\sim}{\delta}^{T}+\underset{\sim}{r}^{T}\right)(\underset{\sim}{Q}+\underset{\sim}{r}) \\ &=\underset{\sim}{\delta}^{T} \underset{\sim}{\delta}+\underset{\sim}{\delta}^{T} \underset{\sim}{r}+\underset{\sim}{\delta} \underset{\sim}{r}^{T}+\underset{\sim}{r}^{T} \underset{\sim}{r} \\ &=\underset{\sim}{\delta}+ \underset{\sim}{\delta} \underset{\sim}{r}+\underset{\sim}{\delta} \underset{\sim}{r}^{T}+\underset{\sim}{r}^{T} \underset{\sim}{r}=\underset{\sim}{\delta}, \end{aligned}

(using the identities $$\underset{\sim}{\delta}^T$$ = $$\underset{\sim}{\delta}$$ and $$\underset{\sim}{\delta}\underset{\sim}{\delta}$$ = $$\underset{\sim}{\delta}$$), and therefore

$\underset{\sim}{r}+\underset{\sim}{r}^{T}+\underset{\sim}{r}^{T} \underset{\sim}{r}=0.\label{eqn:1}$

Now suppose that the rotation is through a very small angle1. In this case, all components of $$\underset{\sim}{r}$$ are $$\ll 1$$, and the third term on the left-hand side of Equation $$\ref{eqn:1}$$ is therefore negligible, leaving us with

$\underset{\sim}{r}+\underset{\sim}{r}^{T}=0.$

So, for an infinitesimal rotation, the rotation matrix equals the identity plus an antisymmetric matrix whose elements are $$\ll 1$$.

1More precisely, we take the limit as the angle goes to zero.